Given a Transformation Matrix $T$, find $T$ relative to a new basis $\beta$

Question

$T(a_1,a_2,a_3) = (3a_1+a_2,a_1+a_3,a_1-a_3)$. $(a_1,a_2,a_3)^T$ is written with regards to the standard basis.

We can figure out $T$ in matrix form by calculating $T(a_1),T(a_2), T(a_3)$. That's easy and we get:

$T$ = $\left(\begin{array}{ccc}3 & 1 & 0 \\1& 0 & 1 \\1 & 0 & -1\end{array}\right)$

$\beta$ is a new basis = $\{(1,0,0), (1,1,0), (1,1,1)\}$.

Here's what I'm trying to find: $T$ relative to this new basis.

I have three questions:

First, How do I methodically set up the problem and solve it?

Second, why does $T*C = S$ work?, where

$T$ is the transformation matrix relative to $(a_1,a_2,a_3)^T$,

$C = \left(\begin{array}{ccc}1 & 1 & 1 \\0& 1 & 1 \\0 & 0 & 1\end{array}\right)$.

and $S$ is the solution = $\left(\begin{array}{ccc}3 & 4 & 4 \\1& 1 & 2 \\1 & 1 & 0\end{array}\right)$.

Third Question: how general is $T*C$ as a solution? Did I just get lucky? It seems all I'm doing is multiplying a new vector (a vector in $\beta$) by an old transformation matrix. Why would this give me a new, correct $T$? Basically, if someone asks me to find $T$ relative to a new basis can I just use $T*C$, where $C$'s columns contain the new basis vectors?

Edit: How do you do this without using inverses?

Answer 1

My linear algebra is a bit rusty, so I think this answer should help you out more than I could: https://math.stackexchange.com/a/340991/221695

In your particular case, you do seem to have gotten lucky. As detailed in the post above, the methodical way is to set up this change-of-base matrix C correctly and then right-multiply your transformation matrix T by the inverse of C.

I may be mistaken, but I think you've treated the basis vectors as row vectors while constructing your C, ie

$$v_1=(1,0,0)$$

and

$$C = \left(\begin{array}{c}v1 \\v2 \\v3\end{array}\right)$$

While it should have been something like:

$$v1 = \left(\begin{array}{c}1 \\0 \\0\end{array}\right)$$

and

$$C=(v1,v2,v3)=\left(\begin{array}{ccc}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 1\end{array}\right)$$

I just tinkered a bit more with this and computing T*C with the C matrix constructed with column vectors as above magically gives the S you specified, which I don't get with your C matrix. Anyway, if you're sure that S is the right answer, I'm not sure the problem is formulated correctly.

Answer 2

Let

$\mathcal{E}=\left\{e_1,e_2,e_3\right\}$

be our canonical base. With this base, transormation T has representation

$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.

Now we have got a new base:

$\mathcal{F}=\left\{e_1,e_1+e_2,e_1+e_2+e_3\right\}$.

Let

$M_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right)$

be the transition between the two bases.

Then canonical coordinates are transormed in new coordinates (with respect to base $\mathcal{F}$ ) by inverse matrix, which is

$N_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{array} \right)$.

Take

$A=\left\{a_1,a_2,a_3\right\}$

and get new coordinates

$B=N_{\mathcal{F}}.A$.

Then, with $S=T.M_{\mathcal{F}}$

we see:

$T.A=T.M_{\mathcal{F}}.N_{\mathcal{F}}.A=S.B$.

It's not a miracle, only lin. Algebra. Key is transformation of basis, which implies transformation of coordinates. That's all.

By the way: Calculating without inverses is not possible. Your transformation with bases must be regular. They must be invertible, otherwise it didn't work.

Let's see. Other basis

$\mathcal{B}=\left\{2 e_1+5 e_3,e_1+e_2+6 e_3,3 e_1+9 e_3\right\}$,

another transition:

$M_{\mathcal{B}}=\left( \begin{array}{ccc} 2 & 1 & 3 \\ 0 & 1 & 0 \\ 5 & 6 & 9 \\ \end{array} \right)$.

The inverse:

$N_{\mathcal{B}}=\left( \begin{array}{ccc} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -\frac{5}{3} & -\frac{7}{3} & \frac{2}{3} \\ \end{array} \right)$.

Old transformation T

$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.

Transformed T:

$S=T.M_{\mathcal{B}}=\left( \begin{array}{ccc} 6 & 4 & 9 \\ 7 & 7 & 12 \\ -3 & -5 & -6 \\ \end{array} \right)$

Transformed A:

$B=N_{\mathcal{B}}.A$.

$T.A=T.M_{\mathcal{B}}.N_{\mathcal{B}}.A=S.B$

Like before.